COERCIVE SOLVABILITY OF PARABOLIC DIFFERENTIAL EQUATIONS WITH DEPENDENT OPERATORS
A. ASHYRALYEV1, A. HANALYEV2 §
Abstract. In the present paper the nonlocal-boundary value problem for the differential equation of parabolic type
v′(t) +A(t)v(t) =f(t) (0≤t≤T), v(0) =v(λ) +φ,0< λ≤T
in an arbitrary Banach space with the linear positive operatorsA(t) is considered. The well-posedness of this problem is established in Banach spacesC0β,γ(E) of all continuous functions E-valued functionsφ(t) on [0, T] satisfying a H¨older condition with a weight (t+τ)γ.New exact estimates in Holder norms for the solution of three nonlocal-boundary value problems for parabolic equations are obtained.
Keywords: Parabolic equations, NBV problems, Banach spaces, positive operators. AMS Subject Classification: 47D06, 35K20, 35M10
1. Introduction. A Cauchy Problem
It is known that (see, e.g., [1]- [5] and the references given therein) many applied prob-lems in fluid mechanics, other areas of physics and mathematical biology were formulated into nonlocal mathematical models. However, such problems were not well investigated in general.
In the paper [6] the well-posedness in the spaces of smooth functions of the nonlocal boundary value problem
v′(t) +Av(t) =f(t)(0≤t≤1), v(0) =v(λ) +µ(0< λ≤1)
for the differential equation in an arbitrary Banach space E with the strongly positive operator A was established. The importance of coercive (well-posedness) inequalities is well-known [10] and [32].
Finally, methods for numerical solutions of the evolution differential equations have been studied extensively by many researchers (see [7]-[9], [11]- [32] and the references therein). Before going to discuss well-posedness of nonlocal-boundary value problem, let us con-sider the abstract Cauchy problem for the differential equation
v′(t) +A(t)v(t) =f(t) (0≤t≤T), v(0) =v0 (1) in an arbitrary Banach spaceE with the linear (unbounded) operatorsA(t). Herev(t) and
f(t) are the unknown and the given functions, respectively, defined on [0, T] with values
1 Department of Mathematics, Fatih University 34500, B¨uy¨uk¸cekmece, Istanbul, Turkey and
Department of Mathematics, ITT University 74400, Ashgabat, Turkmenistan e-mail: [email protected], [email protected]
2 Department of Mathematics, ITT University 74400, Ashgabat, Turkmenistan,
e-mail: [email protected] § Manuscript received 24 January 2012.
TWMS Journal of Applied and Engineering Mathematics Vol.2 No.1 c⃝I¸sık University, Department of Mathematics 2012; all rights reserved.
inE.The derivativev′(t) is understood as the limit in the norm ofEof the corresponding ratio of differences. A(t) is a given, closed, linear operator inEwith domainD(A(t)) =D,
independent oftand dense in E.Finally, v0 is a given element of E.
A function v(t) is called a solution of the problem (1.1) if the following conditions are
satisfied:
i. v(t) is continuously differentiable on the segment [0, T]. The derivative at the endpoints of the segment are understood as the appropriate unilateral derivatives. ii. The element v(t) belongs to D = D(A(t)) for all t ∈ [0, T], and the function
A(t)v(t) is continuous on [0, T].
iii. v(t) satisfies the equation and the initial condition (1.1).
A solution of problem (1.1) defined in this manner will from now on be referred to as a solution of problem (1.1) in the space C(E) = C([0, T], E). Here C(E) stands for the Banach space of all continuous functionsφ(t) defined on [0, T] with values inE equipped with the norm
||φ||C(E)= max
0≤t≤T||φ(t)||E.
From the existence of the such solutions evidently follows thatf(t)∈C(E) andv0 ∈D. We say that the problem (1.1) is well posed in C(E) if the following conditions are satisfied:
1. Problem (1.1) is uniquely solvable for any f(t) ∈ C(E) and any v0 ∈ D. This means that an additive and homogeneous operator v(t) =v(t;f(t), v0) is defined which acts from C(E)×D to C(E) and gives the solution of problem (1.1) in
C(E).
2. v(t;f(t), v0), regarded as an operator fromC(E)×DtoC(E),is continuous. Here
C(E)×D is understood as the normed space of the pairs (f(t), v0), f(t) ∈C(E) and v0∈D,equipped with the norm
||(f(t), v0)||C(E)×D =||f||C(E)+||v0||D.
By Banach’s theorem in C(E) and these properties one has coercive inequality
v′C(E)+∥A(.)v∥C(E)≤MC[∥f∥C(E)+ ||v0||D], (2)
whereMC (1≤MC <+∞) does not depend onv0 and f(t).
The inequality (2) is called the coercivity inequality in C(E) for (1.1). If A(t) = A,
then the coercivity inequality implies the analyticity of the semigroup exp{−sA}(s≥0), i.e. the following estimates
∥exp (−sA)∥E→E,∥sAexp(−sA)∥E→E ≤M(s >0)
hold for someM ∈[1,+∞).Thus, the analyticity of the semigroup exp{−sA}(s≥0) is a necessary for the well-posedness of problem (1.1) inC(E).Unfortunately, the analyticity of the semigroup exp{−sA} (s ≥0) is not a sufficient for the well-posedness of problem (1.1) inC(E).
Suppose that for each t ϵ[0, T] the operator −A(t) generates an analytic semigroup exp{−sA(t)} (s ≥ 0) with exponentially decreasing norm, when s −→ +∞, i.e. the following estimates
hold for some M ∈ [1,+∞), δ ∈ (0,+∞). From this inequality it follows the operator
A−1(t) exists and bounded, and hence A(t) is closed in C(E).
Suppose that the operatorA(t)A−1(s) is H¨older continuous intin the uniform operator topology for each fixeds, that is,
||[A(t)−A(τ)]A−1(s)||E→E ≤M|t−τ|ε,0< ε≤1, (4)
whereM and εare positive constants independent of t, sandτ for 0≤t, s, τ ≤T.
If the functionf(t) is not only continuous, but also continuously differentiable on [0, T],
andv0 ∈D,it is easy to show that the formula
v(t) =v(t,0)v0+
t
∫
0
v(t, s)f(s)ds (5)
gives a solution of problem (1.1). Herev(t, s) is the fundamental solution of (1.1). Now we will give lemmas and estimates from [13] concerning the semigroup exp{−sA(t)} (s ≥ 0) and the fundamental solution v(t, s) of (1.1) and theorem on well-posedness of (1.1) which will be useful in the sequel.
Lemma 1.1. For any0< s < s+τ < T,0≤t≤T and0≤α≤1one has the inequality
∥exp (−sA(t))−exp (−(s+τ)A(t))∥E→E ≤M τ α
(s+τ)α, (6)
where M does not depend on α, t, s, and τ.
Lemma 1.2. For any 0≤s, τ, t≤T and 0≤ε≤1 the following estimates hold:
[exp (−tA(τ))−exp (−sA(τ))]A−1(τ)E→E ≤M|t−s|e−δmin{t,s}, (7)
A(t)[exp (−tA(τ))−exp (−sA(τ))]A−2(τ)E→E ≤M|t−s|e−δmin{t,s}, (8)
where M ≥0 andδ >0 do not depend on ε, t, s, and τ.
Lemma 1.3. For any 0≤s < t≤T and u ϵD the following identities hold:
v(t, s)u=exp{−(t−s)A(s)}u (9)
+
t
∫
s
v(t, z)[A(s)−A(z)]A−1(s)exp {−(z−s)A(s)}A(s)udz,
v(t, s)u=exp{−(t−s)A(t)}u (10)
+
t
∫
s
exp {−(t−z)A(t)}[A(z)−A(t)]v(z, s)udz.
Lemma 1.4. For any 0 ≤ s < t ≤ t+r ≤ T, 0 ≤ α ≤ 1 and 0 ≤ ε ≤ 1 the following estimates hold:
∥v(t, s)∥E→E ≤M, (11)
A(t)v(t, s)A−1(s)E→E ≤M, (12) ∥A(t)v(t, s)∥E→E ≤ M
t−s, (13)
With the help of A(t) we introduce the fractional spaces Eα(E, A(t)), 0 < α < 1,
consisting of allv ϵE for which the following norms are finite: ∥v∥Eα= sup
z>0
z1−α ∥A(t)exp{−zA(t)}v∥E .
From (1.3) and (1.4) it follows that Eα(E, A(t)) = Eα(E, A(0)) for all 0 < α <1 and
0≤t≤T.
A function v(t) is said to be a solution of problem (1.1) in F(E) if it is a solution of this problem inC(E) and the function v′(t) and A(t)v(t) belong toF(E).
As in the case of the space C(E),we say that the problem (1.1) is well-posed inF(E),
if the following two conditions are satisfied:
1. For anyf ∈F(E) andv0 ∈D there exists the unique solutionv(t) =v(t;f(t), v0) inF(E) of problem (1.1).This means that an additive and homogeneous operator
v(t;f(t), v0) is defined which acts from F(E)×D toF(E) and gives the solution of (1.1) in F(E).
2. v(t;f(t), v0), regarded as an operator fromF(E)×DtoF(E),is continuous.Here
F(E)×D is understood as the normed space of the pairs (f(t), v0), f(t) ∈F(E) and v0∈D,equipped with the norm
||(f(t), v0)||F(E)×D =||f||F(E)+||v0||D.
We set F(E) equal to C0β,γ(E),(0≤γ ≤ β,0< β < 1) space, obtained by completion of the set of all smoothE-valued functionsφ(t) on [0, T] with respect to the norm
∥φ∥Cβ,γ
0 (E)
= max
0≤t≤T||φ(t)||E+0≤t<tsup+τ≤T
(t+τ)γ∥φ(t+τ)−φ(t)∥E
τβ .
Let us give, the following theorem on well-posedness of (1.1) inC0β,γ(E) from [13].
Theorem 1.1. Suppose v0′ ∈Eβ−γ, f(t)∈C0β,γ(E)(0≤γ ≤β,0< β <1). Suppose that the assumptions (1.3) and (1.4) hold and 0 < β ≤ ε < 1. Then for the solution v(t) in
C0β,γ(E) of the Cauchy problem (1.1) the coercive inequalities
∥v′∥C(Eβ−γ)≤M[∥v0′ ∥Eβ−γ +β−
1(1−β)−1∥f ∥
C0β,γ(E)], ∥v′ ∥Cβ,γ
0 (E)
+∥A(.)v∥Cβ,γ
0 (E)
≤M[|v0|′ β,γ0 +β−1(1−β)−1 ∥f ∥Cβ,γ
0 (E)]
hold, where M does not depend on β, γ, v′0 and f(t). Here, |w|β,γ0 denotes norm of the Banach spaceE0β,γ consists of those w∈E for which the norm
|w|β,γ0 = max 0≤z≤T||e
−zA(t)w||
E+ sup
0≤z<z+τ≤T
τ−β(z+τ)γ||(e−(z+τ)A(t)−e−zA(t))w||E
is finite.
In the present paper the nonlocal-boundary value problem for differential equation of parabolic type
v′(t) +A(t)v(t) =f(t) (0≤t≤T), v(0) =v(λ) +φ,0< λ≤T (14)
2. Nonlocal Boundary Value Problem. Well-Posedness
Now we will give lemmas on the fundamental solution v(t, s) of (1).
Lemma 2.1. Assume that A(t)A(p)−1=A(t+λ)A(p)−1, p∈[0, T]for any 0≤t≤t+λ
Then, for any 0≤s < t≤t+λ andu ϵD the following identity holds
v(t, s)u=v(t+λ, s)u. (15)
The proof of Lemma 2.1 is based on identities (9) and (10).
Lemma 2.2. Under the assumption of Lemma 2.1 there exists the inverse of the operator
I−v(λ,0)in E and the following estimate holds
(I−v(λ,0))−1
E→E ≤M(λ), (16)
A(0) (I−v(λ,0))−1A(λ)−1
E→E ≤M(λ). (17)
The proof of Lemma 2.2 is based on identity (15).
A function v(t) is called a solution of the problem (14) if the following conditions are satisfied:
i. v(t) is continuously differentiable on the segment [0, T].
ii. The element v(t) belongs to D for all t ∈ [0, T], and the function A(t)v(t) is continuous on [0, T].
iii. v(t) satisfies the equation and the nonlocal boundary condition (14).
We say that the problem (14) is well posed in C(E) if the following conditions are satisfied:
1. Problem (14) is uniquely solvable for anyf(t)∈C(E) and anyφ∈D.This means that an additive and homogeneous operator v(t) = v(t;f(t), φ) is defined which acts fromC(E)×D toC(E) and gives the solution of problem (1.1) inC(E).
2. v(t;f(t), φ), regarded as an operator fromC(E)×DtoC(E),is continuous. Here
C(E)×D is understood as the normed space of the pairs (f(t), φ), f(t) ∈ C(E) and φ∈D,equipped with the norm
||(f(t), φ)||C(E)×D =||f||C(E)+||φ||D.
By Banach’s theorem in C(E) and these properties one has coercive inequality
v′C(E)+∥A(.)v∥C(E)≤MC[∥f∥C(E)+ ||φ||D], (18)
whereMC (1≤MC <+∞) does not depend onφ andf(t).
The inequality (18) is called the coercivity inequality inC(E) for (14). IfA(t) =A,then the coercivity inequality implies the analyticity of the semigroup exp{−sA}(s≥0). Thus, the analyticity of the semigroup exp{−sA}(s≥0) is a necessary for the well-posedness of problem (14) inC(E).Unfortunately, the analyticity of the semigroup exp{−sA} (s≥0) is not a sufficient for the well-posedness of problem (14) inC(E).
A functionv(t) is said to be a solution of problem (14) inF(E) if it is a solution of this problem inC(E) and the functionv′(t) and A(t)v(t) belong toF(E).
As in the case of the space C(E),we say that the problem (14) is well-posed inF(E),
1. For any f ∈F(E) and φ∈D there exists the unique solution v(t) = v(t;f(t), φ) inF(E) of problem (14).This means that an additive and homogeneous operator
v(t;f(t), φ) is defined which acts fromF(E)×D to F(E) and gives the solution of (14) inF(E).
2. v(t;f(t), φ), regarded as an operator fromF(E)×DtoF(E),is continuous.Here
F(E)×D is understood as the normed space of the pairs (f(t), φ), f(t)∈ F(E) and φ∈D,equipped with the norm
||(f(t), φ)||F(E)×D =||f||F(E)+||φ||D.
The main result of present paper is the following theorem on well-posedness of (14) in
C0β,γ(E).
Theorem 2.1. Suppose A(0)φ+f(λ)−f(0) ∈ Eβ−γ, f(t) ∈ C0β,γ(E)(0 ≤ γ ≤ β,0 <
β <1).Suppose that the assumptions (1.3), (1.4) and (15) hold and 0< β≤ε <1. Then for the solutionv(t) in C0β,γ(E) of the nonlocal boundary value problem (14) the coercive inequalities
∥v′∥C(Eβ−γ)≤M(λ)[∥A(0)φ+f(λ)−f(0)∥Eβ−γ +β
−1(1−β)−1∥f ∥
Cβ,γ0 (E)], ∥v′ ∥Cβ,γ
0 (E)+∥A(.)v∥C
β,γ
0 (E)
≤M(λ)[|A(0)φ+f(λ)−f(0)|β,γ0 +β−1(1−β)−1 ∥f ∥Cβ,γ
0 (E)] hold, whereM(λ) does not depend onβ, γ, φ and f(t).
Proof. Ifv(t) is a solution inC0β,γ(E) of problem (14), then it is a solution inC(E) of this problem. Hence, by (5), we get the following representation for the solution of problem (14)
v(t) = v(t,0)v(0) +
∫ t
0
v(t, s)f(s)ds, (19)
v(0) = (I−v(λ,0))−1
(∫ λ
0
v(λ, s)f(s)ds+φ
)
.
Using equation (14) and formula (19), we get
v0′ =v′(0) =−A(0)v(0) +f(0) =−A(0) (I−v(λ,0))−1
(∫ λ
0
v(λ, s)f(s)ds+φ
)
+f(0)
(20)
=A(0)(I −v(λ,0))−1
λ
∫
0
v(λ, s)(f(λ)−f(s))ds
−A(0)(I−v(λ,0))−1
λ
∫
0
v(λ, s)[A(λ)−A(s)]A−1(λ)f(λ)ds
−A(0)(I −v(λ,0))−1((I−v(λ,0))A−1(λ)f(λ) +φ)+f(0)
=A(0)(I −v(λ,0))−1
λ
∫
0
−A(0)(I−v(λ,0))−1
λ
∫
0
v(λ, s)[A(λ)−A(s)]A−1(λ)f(λ)ds
−A(0)A−1(λ)f(λ)−A(0)(I−v(λ,0))−1φ+f(0)
=A(0)(I −v(λ,0))−1
λ
∫
0
v(λ, s)(f(λ)−f(s))ds
−A(0)(I−v(λ,0))−1
λ
∫
0
v(λ, s)[A(λ)−A(s)]A−1(λ)f(λ)ds
+A(0)(I−v(λ,0))−1A−1(0) (−A(0)φ−f(λ) +f(0))
+A(0)(I−v(λ,0))−1 v(λ,0) (A−1(λ)f(λ)−A−1(0)f(0)) +A(0)(I−v(λ,0))−1v(λ,0)A−1(λ) (A(λ)−A(0)) A−1(0)f(λ)
=K1+K2+K3+K4, where
K1 =A(0)(I −v(λ,0))−1
λ
∫
0
v(λ, s)(f(λ)−f(s))ds,
K2 =−A(0)(I −v(λ,0))−1
λ
∫
0
v(λ, s)[A(λ)−A(s)]A−1(λ)f(λ)ds,
K3 =A(0)(I−v(λ,0))−1A−1(0) (−A(0)φ−f(λ) +f(0)),
K4 = A(0)(I−v(λ,0))−1 v(λ,0)
(
A−1(λ)f(λ)−A−1(0)f(0))
+A(0)(I −v(λ,0))−1v(λ,0)A−1(λ) (A(λ)−A(0)) A−1(0) f(λ).
Then the proof of Theorem 2.1 is based on the Theorem 1.1 and the following estimates
∥v0′ ∥Eβ−γ≤M(λ)
[
∥ −A(0)φ−f(λ) +f(0)∥Eβ−γ +β−
1(1−β)−1∥f ∥
C0β,γ(E)
]
, (21)
|v0′|β,γ0 ≤M(λ)
[
| −A(0)φ−f(λ) +f(0)|β,γ0 +β−1(1−β)−1 ∥f ∥Cβ,γ
0 (E) ]
. (22)
Let us estimate Km for any m = 1,2,3,4 in Eβ−γ and E0β,γ, separately. We start with
K1.Applying the inequality (17), we get ∥K1∥Eβ
−γ ≤M(λ)
A(λ)
λ
∫
0
v(λ, s)(f(λ)−f(s))ds
Eβ−γ
, (23)
|K1|β,γ0 ≤M(λ)
A(λ)
λ
∫
0
v(λ, s)(f(λ)−f(s))ds
β,γ
0
Using estimates (3), (11), (12) and (15), we obtain
z1−β+γ∥A(λ)exp{−zA(λ)}A(λ)
λ
∫
0
v(λ, s)(f(λ)−f(s))ds||E (25)
≤z1−β+γ λ
∫
0
||A2(λ) exp{−zA(λ)}v(t, s)||E→E||f(λ)−f(s)||Eds
≤M z1−β+γ λ
∫
0 min
[
1
z2, 1 (λ−s)2
]
(λ−s)βλ−γds||f||Cβ,γ
0 (E)
≤M1z1−β+γ
λ
∫
0
(λ−s)βds
(z+λ−s)2λγ||f||C0β,γ(E) for allz >0 . We will prove that
z1−β+γ λ
∫
0
(λ−s)βds
(z+λ−s)2λγ ≤
1
β(1−β) (26)
for anyz >0.If z≤λ, then
z1−β+γ λ
∫
0
(λ−s)βds
(z+λ−s)2λγ ≤z
1−β λ
∫
0
ds
(z+λ−s)2−β ≤
1 1−β.
Ifλ≤z, then
z1−β+γ λ
∫
0
(λ−s)βds
(z+λ−s)2λγ ≤
1
zβ−γλγ λ
∫
0
ds
(λ−s)1−β = λβ−γ βzβ−γ <
1
β.
From these estimates it follows (26).Applying (26),(23),(25), we get
∥K1∥Eβ
−γ ≤
M(λ)
β(1−β)||f||Cβ,γ0 (E). (27)
Using estimates (3), (11), (12) and (15), we obtain
∥exp{−zA(λ)}A(λ)
λ
∫
0
v(λ, s)(f(λ)−f(s))ds||E (28)
≤
λ
∫
0
||A(λ) exp{−zA(λ)}v(t, s)||E→E||f(λ)−f(s)||Eds
≤M λ
∫
0 min
[
1
z,
1 (λ−s)
]
(λ−s)βλ−γds||f||Cβ,γ
≤M1
λ
∫
0 1
z+λ−s(λ−s)
βλ−γds||f||
C0β,γ(E)≤M1
λ
∫
0
ds
(λ−s)1−βλ−
γds||f|| C0β,γ(E) ≤ M1
β λ
β−γ||f||
C0β,γ(E) ≤
M1
β T
β−γ||f|| C0β,γ(E)
for anyz >0. Ifλ≤τ +z, then Using estimates (6) forα=β, (11), (12) and (15), we obtain
τ−β(z+τ)γ
(e−(z+τ)A(λ)−e−zA(λ))A(λ)
λ
∫
0
v(λ, s)(f(λ)−f(s))ds
E
(29)
≤M τ−β(z+τ)γ τ
β
(z+τ)β
A(λ)
λ
∫
0
v(λ, s)(f(λ)−f(s))ds
E
≤ M1 (z+τ)β−γ
λ
∫
0
ds
(λ−s)1−βλ− γ||f||
Cβ,γ0 (E)≤
M1 (z+τ)β−γ
λβ−γ
β ||f||C0β,γ(E)≤
M2
β ||f||C0β,γ(E) for any 0≤z < z+τ ≤T. Ifτ+z≤λ andτ ≤z, then using estimates (12) and (3), we obtain
τ−β(z+τ)γ
(e−(z+τ)A(λ)−e−zA(λ))A(λ)
λ
∫
0
v(λ, s)(f(λ)−f(s))ds
E
(30)
≤M τ−β(z+τ)γ
λ
∫
0
A(λ)(e−(z+τ)A(λ)−e−zA(λ))v(λ, s)(f(λ)−f(s))
Eds
≤ M1 (z+τ)−γ
λ
∫
0
τ1−βds
(z+λ−s)2−βλ− γ||f||
C0β,γ(E)≤
M2
z1−β τ1−β
1−β||f||C0β,γ(E) ≤
M3
1−β||f||C0β,γ(E) for any 0≤z < z+τ ≤T. If τ+z≤λ and τ ≥z, then, using estimates (3), (11), (12) and (8), we obtain
τ−β(z+τ)γ
(e−(z+τ)A(λ)−e−zA(λ))A(λ)
λ
∫
0
v(λ, s)(f(λ)−f(s))ds
E
(31)
≤M τ−β(z+τ)γ
λ∫−τ
0
A(λ)(e−(z+τ)A(λ)−e−zA(λ))v(λ, s)(f(λ)−f(s))
Eds
+M τ−β(z+τ)γ
λ
∫
λ−τ
A(λ)(e−(z+τ)A(λ)−e−zA(λ))v(λ, s)(f(λ)−f(s))
Eds
≤ τ−βM1 (z+τ)−γ
λ∫−τ
0
τ ds
(z+λ−s)2−βλ− γ||f||
+ τ
−βM
1 (z+τ)−γ
λ
∫
λ−τ
ds
(λ−s)1−βλ− γ||f||
Cβ,γ0 (E)≤
M2 (z+τ)β
τ
(1−β) (z+τ)1−β||f||C0β,γ(E)
+ M2 (z+τ)β
τβ
β ||f||C0β,γ(E)≤
M3
β(1−β)||f||C0β,γ(E) for any 0≤z < z+τ ≤T. Applying (24),(28),(29), (30),(31),we get
|K1|β,γ0 ≤ M(λ)
β(1−β)||f||C0β,γ(E). (32)
Now, we estimateK2.Applying the inequality (17), we get ∥K2∥Eβ
−γ ≤M(λ)
A(λ)
λ
∫
0
v(λ, s)[A(λ)−A(s)]A−1(λ)f(λ)ds
Eβ−γ
, (33)
|K2|β,γ0 ≤M(λ)
A(λ)
λ
∫
0
v(λ, s)[A(λ)−A(s)]A−1(λ)f(λ)ds
β,γ 0 . (34)
Using estimates (3), (11), (12) and (15), we obtain
z1−(β−γ)
A(λ) exp{−zA(λ)} λ
∫
0
A(λ)v(λ, s)[A(λ)−A(s)]A−1(λ)f(λ)ds
E
(35)
≤z1−β+γ λ
∫
0
||A2(λ) exp{−zA(λ)}v(λ, s)||E→E||[A(λ)−A(s)]A−
1(λ)||
E→E||f(λ)||Eds
≤z1−β+γ λ ∫ 0 min [ 1
z2, 1 (λ−s)2
]
(λ−s)εds||f||Cβ,γ
0 (E) ≤
≤M1z1−β+γ
λ
∫
0
(λ−s)εds
(z+λ−s)2||f||C0β,γ(E) for allz >0 . We will prove that
z1−β+γ λ
∫
0
(λ−s)εds
(z+λ−s)2 ≤
M
β(1−β) (36)
for anyz >0.Ifz≤λ, then
z1−β+γ λ
∫
0
(λ−s)εds
(z+λ−s)2 ≤z 1−βλγ
λ
∫
0
(z+λ)ε−βds
(z+λ−s)2−β ≤
2λε−β+γ
1−β ≤
2T2 1−β.
z1−β+γ λ
∫
0
(λ−s)εds
(z+λ−s)2 ≤ 1
zβ−γ λ
∫
0
ds
(λ−s)1−ε = λε εzβ−γ <
λε ελβ−γ ≤
T2
β .
From these estimates it follows (36).Applying (36),(33),(35), we get
∥K2∥Eβ
−γ ≤
M(λ)
β(1−β)||f||Cβ,γ0 (E). (37)
Using estimates (3), (11), (12) and (15), we obtain
∥exp{−zA(λ)}A(λ)
λ
∫
0
v(λ, s)[A(λ)−A(s)]A−1(λ)f(λ)ds||E (38)
≤
λ
∫
0
||A(λ) exp{−zA(λ)}v(t, s)||E→E[A(λ)−A(s)]A−1(λ)E→E||f(λ)||Eds
≤M λ ∫ 0 min [ 1 z, 1 (λ−s)
]
(λ−s)εds||f||Cβ,γ
0 (E)
≤M1
λ
∫
0
1
z+λ−s(λ−s)
εds||f||
C0β,γ(E)≤M1
λ
∫
0
ds
(λ−s)1−εds||f||C0β,γ(E) ≤ M1
ε λ
ε||f||
C0β,γ(E) ≤
M1
ε T
ε||f||
C0β,γ(E)≤
M1
β T||f||C0β,γ(E)
for anyz >0. Ifλ≤τ+z, then Using estimates (3), (6) forα=β, (11), (12) and (15), we obtain
τ−β(z+τ)γ
(e−(z+τ)A(λ)−e−zA(λ))A(λ)
λ
∫
0
v(λ, s)[A(λ)−A(s)]A−1(λ)f(λ)ds
E
(39)
≤M τ−β(z+τ)γ τ
β
(z+τ)β
A(λ)
λ
∫
0
v(λ, s)[A(λ)−A(s)]A−1(λ)f(λ)ds
E
≤ M1 (z+τ)β−γ
λ
∫
0
ds
(λ−s)1−ε||f||C0β,γ(E)≤
M1 (z+τ)β−γ
λε
ε ||f||C0β,γ(E)≤
M2T2
β ||f||C0β,γ(E) for any 0≤z < z+τ ≤T. If τ +z≤λ and τ ≤z, then using estimates (3), (11) and (13), we obtain
τ−β(z+τ)γ
(e−(z+τ)A(λ)−e−zA(λ))A(λ)
λ
∫
0
v(λ, s)[A(λ)−A(s)]A−1(λ)f(λ)ds
E
(40)
≤M τ−β(z+τ)γ
λ
∫
0
A(λ)(e−(z+τ)A(λ)−e−zA(λ))v(λ, s)[A(λ)−A(s)]A−1(λ)f(λ)
≤ M1 (z+τ)−γ
λ
∫
0
(z+λ)ε−βτ1−βds
(z+λ−s)2−β ||f||C0β,γ(E) ≤
M2
z1−β
τ1−βTε−β
1−β ||f||C0β,γ(E)≤
M3
1−β||f||C0β,γ(E) for any 0≤z < z+τ ≤T. If τ +z≤λ and τ ≥z, then using estimates (12) and (3), we obtain
τ−β(z+τ)γ
(e−(z+τ)A(λ)−e−zA(λ))A(λ)
λ
∫
0
v(λ, s)[A(λ)−A(s)]A−1(λ)f(λ)ds
E
(41)
≤M τ−β(z+τ)γ
λ−τ
∫
0
A(λ)(e−(z+τ)A(λ)−e−zA(λ))v(λ, s)[A(λ)−A(s)]A−1(λ)f(λ)
Eds
+M τ−β(z+τ)γ
λ
∫
λ−τ
A(λ)(e−(z+τ)A(λ)−e−zA(λ))v(λ, s)[A(λ)−A(s)]A−1(λ)f(λ)
Eds
≤ τ−βM1 (z+τ)−γ
λ∫−τ
0
λε−βτ ds
(z+λ−s)2−β||f||C0β,γ(E)
+ τ
−βM
1 (z+τ)−γ
λ
∫
λ−τ
ds
(λ−s)1−ε||f||C0β,γ(E)≤
M2Tε−β+γτ1−β
(1−β) (z+τ)1−β||f||Cβ,γ0 (E) + M2
(z+τ)β−γ τε
ε ||f||C0β,γ(E)≤
M3
β(1−β)||f||C0β,γ(E) for any 0≤z < z+τ ≤T. Applying (38),(39),(40), (41),we get
|K2|β,γ0 ≤ M(λ)
β(1−β)||f||C0β,γ(E). (42)
Applying the inequality (17), we get ∥K3∥Eβ
−γ ≤M(λ)∥A(0)φ+f(λ)−f(0)∥Eβ−γ, (43)
|K3|β,γ0 ≤M(λ)|A(0)φ+f(λ)−f(0)|β,γ0 . (44) Finally, we estimateK4.Applying the inequality (17), we get
∥K4∥Eβ
−γ ≤M(λ)∥A(λ)v(λ,0)
×( (A−1(λ)f(λ)−A−1(0)f(0))+A−1(λ) (A(λ)−A(0)) A−1(0) f(λ))Eβ
−γ, (45)
|K4|β,γ0 ≤M(λ)|A(λ)v(λ,0)
×( (A−1(λ)f(λ)−A−1(0)f(0))+A−1(λ) (A(λ)−A(0)) A−1(0)f(λ))β,γ0 . (46) Applying the inequality (17), we get
z1−(β−γ)A2(λ) exp{−zA(λ)}v(λ,0)
×( (A−1(λ)f(λ)−A−1(0)f(0))+A−1(λ) (A(λ)−A(0)) A−1(0)f(λ))E (47)
+||[A(λ)−A(0)]A−1(0)||E→E||f(λ)||E)
≤M z1−β+γmin
[
1
z,
1
λ
]
||f||Cβ,γ
0 (E)≤M1
z1−β+γ
z+λ ||f||C0β,γ(E)≤M1λ
−β+γ||f|| Cβ,γ0 (E)
for allz >0. Applying (47),(45), we get ∥K4∥Eβ
−γ ≤M1(λ)||f||C0β,γ(E). (48)
Using estimates (3), (11), (12) and (15), we obtain ∥exp{−zA(λ)}A(λ)v(λ,0)
×( (A−1(λ)f(λ)−A−1(0)f(0))+A−1(λ) (A(λ)−A(0)) A−1(0)f(λ))E (49) ≤ ||A(λ) exp{−zA(λ)}v(λ,0)A−1(λ)||E→E(||f(λ)||E+||A(λ)A−1(0)||E→E||f(0)||E
+||[A(λ)−A(0)]A−1(0)||E→E||f(λ)||E
)
≤Mmin
[
1
z,
1
λ
]
||f||Cβ,γ
0 (E)≤M1
1
λ||f||C0β,γ(E)
for anyz >0. Ifλ≤τ +z, then Using estimates (6) forα=β, (11), (12) and (15), we obtain
τ−β(z+τ)γ(e−(z+τ)A(λ)−e−zA(λ))A(λ)v(λ,0) (50) ×( (A−1(λ)f(λ)−A−1(0)f(0))+A−1(λ) (A(λ)−A(0)) A−1(0) f(λ))E
≤M τ−β(z+τ)γ τ
β
λ(z+τ)β
( (A−1(λ)f(λ)−A−1(0)f(0))+A−1(λ) (A(λ)−A(0)) A−1(0) f(λ))
E
≤ M1
λ(z+τ)β−γ
(
λβ−γ+λε
)
||f||Cβ,γ
0 (E) ≤M2(λ)||f||C
β,γ
0 (E)
for any 0≤z < z+τ ≤T. Ifτ +z≤λ and τ ≤z, then using estimates (12) and (3), we obtain
τ−β(z+τ)γ(e−(z+τ)A(λ)−e−zA(λ))A(λ)v(λ,0) (51) ×( (A−1(λ)f(λ)−A−1(0)f(0))+A−1(λ) (A(λ)−A(0)) A−1(0) f(λ))E
≤M(z+τ)γ
(
τ1−β
(z+λ)2+γ−β +
τ1−β
(z+λ)2−ε
)
||f||Cβ,γ
0 (E)≤
M1(λ)||f||Cβ,γ
0 (E)
for any 0≤z < z+τ ≤T. If τ +z≤λ and τ ≥z, then using estimates (12) and (3), we obtain
τ−β(z+τ)γ(e−(z+τ)A(λ)−e−zA(λ))A(λ)v(λ,0) (52) ×( (A−1(λ)f(λ)−A−1(0)f(0))+A−1(λ) (A(λ)−A(0)) A−1(0) f(λ))E
≤ M1 (z+τ)−γ
(
τ1−β
(z+λ)2+γ−β +
τ1−β
(z+λ)2−ε
)
||f||Cβ,γ
0 (E)≤M1(λ)||f||C
β,γ
0 (E)
for any 0≤z < z+τ ≤T. Applying (46),(49),(50), (51),(52),we get
|K4|β,γ0 ≤M1(λ)||f||Cβ,γ
0 (E). (53)
Combining the estimates (27), (37), (43), (48) and (32), (42), (44), (53),we get estimates
It is easy to show that
|u|β,γ0 ≤ M
β−γ||u||Eβ−γ(u∈Eβ−γ). (54)
Theorem 2.1 admit the following corollary.
Theorem 2.2. Suppose A(0)φ+f(λ)−f(0) ∈ Eβ−γ, f(t) ∈ C0β,γ(E)(0 ≤ γ ≤ β,0 < β <1).Suppose that the assumptions (1.3), (1.4) and (15) hold and 0< β≤ε <1. Then for the solutionv(t) in C0β,γ(E) of the nonlocal boundary value problem (14) the coercive inequalities
∥v′ ∥Cβ,γ
0 (E)+∥A(.)v∥C
β,γ
0 (E)+∥v ′ ∥
C(Eβ−γ)
≤M(λ)[ 1
β−γ||A(0)φ+f(λ)−f(0)||Eβ−γ +β
−1(1−β)−1 ∥f ∥
C0β,γ(E)]
hold, whereM(λ) does not depend onβ, γ, φ and f(t).
3. Applications
First, we consider the nonlocal boundary value problem for parabolic equation
∂u
∂t −a(t, x) ∂2u
∂x2 +δu=f(t, x),0< t < T,0< x <1, (55)
u(0, x) =u(λ, x) +φ(x), 0≤x≤1,
u(t,0) =u(t,1), ux(t,0) =ux(t,1), 0≤t≤T,
wherea(t, x), φ(x) and f(t, x) are given sufficiently smooth functions anda(t, x) =a(t+
λ, x)>0, δ >0 is a sufficiently large number.
We introduce the Banach spaces Cβ[0,1] (0< β <1) of all continuous functions φ(x) satisfying a H¨older condition for which the following norms are finite
∥φ∥Cβ[0,1]=∥φ∥C[0,1]+ sup
0≤x<x+τ≤1
|φ(x+τ)−φ(x)|
τβ ,
where C[0,1] is the space of the all continuous functions φ(x) defined on [0,1] with the usual norm
∥φ∥C[0,1]= max
0≤x≤1|φ(x)|. It is known that the differential expression
At,xv=−a(t, x)v′′(t, x) +δv(t, x)
define a positive operatorAt,x acting inCβ[0,1] with domainCβ+2[0,1] and satisfying the conditionsv(0) =v(1), vx(0) =vx(1).
Therefore, we can replace the nonlocal boundary value problem (55) by the abstract nonlocal boundary value problem (14). We can obtain that
Theorem 3.1. For the solution of nonlocal boundary value problem (55) the following coercive inequaly is valid:
∥u∥C1+β,γ
0 (Cµ[0,1])+∥u∥C
β,γ
0 (C2+µ[0,1]) +∥u∥C(C2(β−γ)+µ[0,1])
≤ M(λ, µ)
β(1−β) ∥f ∥C0β,γ(Cµ[0,1])+
M(λ, µ)
β−γ ||−a(0,·) ∂2φ(·)
∂x2 +δφ(·)+f(λ,·)−f(0,·)||C2(β−γ)+µ[0,1],
0<2(β−γ) +µ <1, 0≤γ ≤β,0≤µ≤1.
The proof of Theorem 3.1 is based on the abstract Theorem 2.1 and on the following theorem on the structure of the fractional spacesEα(C[0,1], At,x).
Theorem 3.2. Eα(C[0,1], At,x) =C2α[0,1] for all0< α < 12,0≤t≤T [33].
Second, let Ω be the unit open cube in the n-dimensional Euclidean space Rn (0 < xk<1,1≤k≤n) with boundary S, Ω = Ω∪S. In [0, T]× Ω we consider the nonlocal
boundary value problem for the multidimensional parabolic equation
∂u(t, x)
∂t −
n
∑
r=1
αr(t, x)∂
2u(t, x)
∂x2
r
+δu(t, x) =f(t, x), (56)
−
n
∑
r=1
αr(t, x)
∂2φ(x)
∂x2
r
+δφ(x) +f(λ, x)−f(0, x) = 0, x= (x1, . . . , xn)∈Ω,0< t < T, u(0, x) =u(λ, x) +φ(x), x∈Ω,
u(t, x) = 0, x∈S,
where αr(t, x), f(t, x) (t ∈ [0, T], x ∈ Ω), φ(x)(x ∈ Ω) are given smooth functions and
αr(t, x) =αr(t+λ, x)>0, δ >0 is a sufficiently large number.
We introduce the Banach spaces C01β (Ω) (β = (β1,· · ·, βn),0 < xk <1, k = 1, . . . , n)
of all continuous functions satisfying a H¨older condition with the indicator β = (β1,· · ·, βn), βk∈(0,1),1≤k≤nand with weightxβkk (1−xk−hk)βk,0≤xk < xk+hk≤1,1≤ k≤n which equipped with the norm
∥f ∥Cβ
01(Ω)=∥f ∥C(Ω)
+ sup
0≤xk<xk+hk≤1,1≤k≤n|
f(x1, . . . , xn)−f(x1+h1, . . . , xn+hn)|
n
∏
k=1
h−kβkxβkk (1−xk−hk)βk,
whereC(Ω)-is the space of the all continuous functions defined on Ω,equipped with the norm
∥f ∥C(Ω)= max
x∈Ω|
f(x)|.
It is known that the differential expression
At,xv=−
n
∑
r=1
αr(t, x)
∂2v(t, x)
∂x2 +δv(t, x)
defines a positive operator At,x acting on C01β(Ω) with domain D(At,x) ⊂ C012+β(Ω) and satisfying the conditionv = 0 onS.
Therefore, we can replace the nonlocal boundary value problem (56) by the abstract nonlocal boundary value problem (14). We can obtain that
Theorem 3.3. For the solution of the nonlocal boundary value problem (56) the following coercive inequality is valid:
∥u∥C1+β,γ
0 (C
µ
01(Ω)) +
n
∑
r=1 ∥ ∂2u
∂x2
r
∥Cβ,γ
0 (C
µ
01(Ω))
≤ M(µ)
where M(µ) is independent ofβ, γ and f(t, x), φ(x).
The proof of Theorem 3.3 is based on the abstract Theorems 2.1, the coercivity inequal-ity for an elliptic operator At,x inC01µ(Ω).
Third, we consider the nonlocal boundary value problem on the range {0≤t≤T, x∈
Rn}for the 2m-th order multidimensional parabolic equation ∂u
∂t +
∑
|r|=2m
ar(t, x)
∂|τ|u ∂xr1
1 · · ·∂xrnn
+δu(t, x) =f(t, x), (57)
0< t < T, x, r∈Rn,|r|=r1+· · ·+rn, u(0, x) =u(λ, x) +φ(x), x∈Rn,
wherear(t, x) =ar(t+λ, x) and f(t, x), φ(x) are given sufficiently smooth functions and
δ >0 is the sufficiently large number .
Let us consider a differential operator with constant coefficients of the form
B= ∑
|r|=2m br ∂
r1+...+rn
∂xr1
1 · · ·∂x
rn n
,
acting on functions defined on the entire spaceRn.Herer ∈Rnis a vector with
nonnega-tive integer components, |r|=r1+· · ·+rn.Ifφ(y) (y= (y1,· · ·, yn)∈Rn) is an infinitely differentiable function that decays at infinity together with all its derivatives, then by means of the Fourier transformation one establishes the equality
F(Bφ) (ξ) =B(ξ)F(φ) (ξ).
Here the Fourier transform operator is defined by the rule
F(φ) (ξ) = (2π)−n/2
∫
Rn
exp{−i(y, ξ)}φ(y)dy,
(y, ξ) =y1ξ1+· · ·+ynξn.
The functionB(ξ) is called the symbol of the operatorBand is given by
B(ξ) = ∑
|r|=2m
br(iξ1)r1 · · ·(iξn)rn.
We will assume that the symbol
Bt,x(ξ) = ∑
|r|=2m
ar(t, x) (iξ1)r1· · ·(iξn)rn, ξ = (ξ1,· · ·, ξn)∈Rn
of the differential operator of the form
Bt,x= ∑
|r|=2m
ar(t, x) ∂
|r|
∂xr1
1 · · ·∂x
rn n
(58)
acting on functions defined on the spaceRn, satisfies the inequalities
0< M1|ξ|2m≤(−1)mBt,x(ξ)≤M2|ξ|2m <∞
forξ ̸= 0. The problem (57) has a unique smooth solution. This allows us to reduce the nonlocal boundary value problem (57) by the abstract nonlocal boundary value problem (14) in a Banach space E = Cµ(Rn) of all continuous bounded functions defined on Rnsatisfying a H¨older condition with the indicator µ ∈ (0,1) with a strongly positive
Theorem 3.4. For the solution of the nonlocal boundary value problem (57) the following coercivity inequality is satisfied
∥u∥C1+β,γ
0 (Cµ(Rn))
+ ∑
|τ|=2m
∥ ∂|r|u
∂xr1
1 · · ·∂x
rn n ∥C
β,γ
0 (Cµ(Rn))
+∥u∥C(C2(β−γ)+µ(Rn))≤
M(λ, µ)
β(1−β) ∥f ∥C0β,γ(Cµ[0,1])
+M(λ, µ)
β−γ ||
∑
|r|=2m
ar(0,·) ∂
|τ|φ(·)
∂xr1
1 · · ·∂x
rn n
+δφ(·) +f(λ,·)−f(0,·)||C2(β−γ)+µ(Rn),
0<2(β−γ) +µ <1, 0≤γ ≤β,0≤µ≤1.
Here M(λ, µ) is independent of γ, β, f(t, x), φ(x).
The proof of Theorem 3.4 is based on the abstract Theorems 2.1, the coercivity inequal-ity for an elliptic operatorAt,x inCµ(Rn) and on the following theorem on the structure of the fractional spacesEα(Cµ(Rn), At,x).
Theorem 3.5. Eα(Cµ(Rn), At,x) =C2mα+µ(Rn)for all 0< α < 21m and0≤t≤T [10]. Acknowledgements: The authors would like to thank Prof. Pavel Sobolevskii (Jerusalem, Israel), for his helpful suggestions to the improvement of this paper.
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Anal. Theo. Meth. and Appl., 30, (4), 1915-1926.
Allaberen Ashyralyev is a full professor in the Department of Mathe-matics at the Fatih University, Istanbul, Turkey and is a joint professor in International Turkmen-Turk University, Ashgabat, Turkmenistan. He com-pleted his first and second PhD Degrees (candidate and doctor of sciences in Mathematics) from Functional Analysis and Operator Equations Depart-ment of Russia Voronezh State University (1983) and Mathematics Institute of Ukraine Science Academy (1992), respectively. His research field is the theory of ordinary and partial differential equations, stochastic partial differ-ential equations numerical analysis, computational mathematics, numerical functional analysis and their applications. He has been the member of the advisory board of a number of national and international mathematics con-ferences and workshops. He is author of more than sixty of articles published in international ISI journals and two monographs published by Birkhauser-Verlag, in Operator Theory: Advances and Applications.
